作业帮试求正整数k,使f(x)=sinkx*sin^k(x)+coskx*cos^k(x)—cos^k(2x)的值不依赖于x
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试求正整数k,使f(x)=sinkx*sin^k(x)+coskx*cos^k(x)—cos^k(2x)的值不依赖于x
试求正整数k,使f(x)=sinkx*sin^k(x)+coskx*cos^k(x)—cos^k(2x)的值不依赖于x
试求正整数k,使f(x)=sinkx*sin^k(x)+coskx*cos^k(x)—cos^k(2x)的值不依赖于x
代入x = 0可得f(0) = 0.
代入x = π/k,可得f(π/k) = -cos(π/k)^k-cos(2π/k)^k.
若f(x)不依赖于x,则有-cos(π/k)^k-cos(2π/k)^k = f(π/k) = f(0) = 0.
若k为偶数,有-cos(π/k)^k ≤ 0且-cos(2π/k)^k ≤ 0.
而由cos(2π/k) = 2cos²(π/k)-1,cos(π/k)与cos(2π/k)不能同时为0.
于是-cos(π/k)^k-cos(2π/k)^k < 0,矛盾.
因此k必须为奇数.
则由-cos(π/k)^k-cos(2π/k)^k = 0,有cos(π/k) = -cos(2π/k) = 1-2cos²(π/k).
解得cos(π/k) = -1,1/2.
而k是正整数,故π/k∈(0,π],于是有π/k = π,π/3,即k = 1,3.
将k = 1代回得f(x) = sin²(x)+cos²(x)-cos²(2x) = 1-cos²(2x)依赖于x,故舍去.
而将k = 3代回得f(x) = sin(3x)sin³(x)+cos(3x)cos³(x)-cos³(2x)
= (3sin(x)-4sin³(x))sin³(x)+(4cos³(x)-3cos(x))cos³(x)-(cos²(x)-sin²(x))³
= -3(cos(x)^4-sin(x)^4)+4(cos(x)^6-sin(x)^6)-(cos²(x)-sin²(x))³
= (cos²(x)-sin²(x))(-3(cos²(x)+sin²(x))+4(cos(x)^4+cos²(x)sin²(x)+sin(x)^4)-(cos²(x)-sin²(x))²)
= (cos²(x)-sin²(x))(-3+3cos(x)^4+6cos²(x)sin²(x)+3sin(x)^4)
= (cos²(x)-sin²(x))(-3+3(cos²(x)+sin²(x))²)
= 0.
即f(x)确实不依赖于x,k = 3满足要求,是问题的解.